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Ierofanga [76]
3 years ago
13

Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c

ontains he, and flask c contains h2. rank the flask contents in terms of:
Chemistry
1 answer:
Margaret [11]3 years ago
6 0

First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

∴ Pressure :

 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas 

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g

∴ the rank of diffusion:

A > B > C

D) The rank of the Total kinetic energy of the molecules:

when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 

∵ Mw A < Mw B < Mw C 

∴no .of molecules of A > B >C

∴ the rank of total kinetic energy is:

A > B > C

e) the rank of density:

when ρ = m/ v 

and m is the mass & v is the volume and we have both is the same for A, B, and C

so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C 

∴the rank of the collision frequency is:

A > B > C 

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7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat
Pie

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

<h3>9.69g of S₂Cl₂</h3>
3 0
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URGENT:<br> Do you switch charges for ionic or covalent bonds when naming them?
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ionic

Explanation:

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If the temperature of water increases as you heat it, the temperature is the independent variable.
Simora [160]
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A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is
liubo4ka [24]
The spacing is 14.4 nm (nanometers).
It is 14.4 * 10^(-9) m
And 1 mm = 1 * 10^(-3) m
14.4 nm * 105 = 1,512 nm = 1.512 micrometers = 0.001512 mm
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3 years ago
Calculate the concentration of H3O⁺ in a solution that contains 5.5 × 10-5 M OH⁻ at 25°C. Identify the solution as acidic, basic
harina [27]

Answer : The correct option is, (A) 1.8\times 10^{-10}M, basic.

Explanation : Given,

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Now we have to calculate the pH.

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Now we have to calculate the H_3O^+ concentration.

pH=-\log [H_3O^+]

9.74=-\log [H_3O^+]

[H_3O^+]=1.8\times 10^{-10}M

As we know that, when the pH value is less than 7 then the solution acidic in nature and when the pH value is more than 7 then the solution basic in nature.

From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.

Therefore, the H_3O^+ concentration is, 1.8\times 10^{-10}M, basic.

3 0
3 years ago
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